Question

tìm x,y biết |x^2+2x|+|y^2-9|=0

giải cụ thể từng bước cho mik nha mik cảm ơn trước

Answer 1

Vì \(\left|x^2+2x\right|\ge0;\left|y^2-9\right|\ge0\)

Dấu ''='' xảy ra <=> \(x^2+2x=0\Leftrightarrow x\left(x+2\right)=0\Leftrightarrow x=0;x=-2\)

\(y^2-9=0\Leftrightarrow\left(y-3\right)\left(y+3\right)=0\Leftrightarrow y=\pm3\)

Answer 2

Ta có :

$\mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid =0$

Do {|x2+2x|≥0|y2−9|≥0{|x2+2x|≥0|y2−9|≥0

$\to \mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid \ge 0$

Mà $\mid \mid {}^{}+2x\mid \mid +\mid \mid {}^{}-9\mid \mid =0$

$\to$ {|x2+2x|=0|y2−9|=0{|x2+2x|=0|y2−9|=0

$\to$ $\{\begin{array}{c}{}^{}+2x=0\\ {}^{}-9=0\end{array}$

$\to$ $\{\begin{array}{c}x(x+2)=0\\ {}^{}=9\end{array}$

$\to$ $⎧⎪\; ⎪\; ⎪\; ⎪⎨⎪\; ⎪\; ⎪\; ⎪⎩\begin{array}{c}[\begin{array}{c}x=0\\ x+2=0\end{array}\\ [\begin{array}{c}y=3\\ y=-3\end{array}\end{array}$

$\to$ $⎧⎪\; ⎪\; ⎪\; ⎪⎨⎪\; ⎪\; ⎪\; ⎪⎩\begin{array}{c}[\begin{array}{c}x=0\\ x=-2\end{array}\\ [\begin{array}{c}y=3\\ y=-3\end{array}\end{array}$

Vậy $x,y\in \{0;3\};\{0;-3\};\{-2;3\};\{-2;-3\}$